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Finite Square-Well Potential |
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| Aug28-10, 04:01 PM | #1 |
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Finite Square-Well Potential
24. Apply the boundary conditions to the finite square-well potential at x=0 to find the relationships between the coefficients A, C, and D and the ratio C/D.
I understand the wave equations in the three separate regions. For this question I need to only consider I, II. The wave equations need to decrease to zero as x approaches positive or negative infinity. The wave equation and its derivative need to be continuous as well. Thus, the wave equation of I equals II. My professor did a similar problem last semester, but I can't make sense of his procedure. I think the delta function is in it, etc. http://i111.photobucket.com/albums/n...g?t=1283054054 |
| Aug28-10, 08:57 PM | #2 |
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Your professor wrote the two equations that correspond to
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| Aug28-10, 10:59 PM | #3 |
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I also re-worked my part, too. http://i111.photobucket.com/albums/n...g?t=1283054054 He has 1 = A + B K = ik (A - B ik/K = (A + B)/(A - B) (ik + K)/(ik - K) = A/B = delta I understand the manipulation up until here. I still don't know how in the heck this helps me related A, C, and D. Do I do the same thing? |
| Aug28-10, 11:56 PM | #4 |
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Finite Square-Well Potential
I think you're getting your coefficients mixed up. Your equations should be
A = C+D αA = ik(C-D) I'm not sure what your professor is doing. It looks like his A and B are your C and D and his K is your alpha. He took (your) A to be equal to 1 for some reason. The delta is not the delta function. It's just the quantity which equals A/B. |
| Aug29-10, 12:07 AM | #5 |
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Here's what the professor did last week. I assume the book is looking for something like this. I don't know how he got this. http://i111.photobucket.com/albums/n...g?t=1283058414 |
| Aug29-10, 12:44 AM | #6 |
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You have two equations and three unknowns, so you can solve for two of them, say C and D, in terms of the other, A.
That's what your professor did except in his case, there were four unknowns, so he solved for B and C in terms of A and D. |
| Aug29-10, 04:19 PM | #7 |
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Oh, for the 1 = A + B, I think he used the free particle solution for the wave heading from the negative x direction towards the potential. |
| Aug31-10, 09:15 PM | #8 |
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| Aug31-10, 11:59 PM | #9 |
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Rewriting the equations a bit, you get
[tex]\begin{align*} B - C & = -A + D \\ k_0 B + kC &= k_0 A + kD \end{align*} [/tex] Multiply the first equation by k, add it the second, and solve for B. |
| Sep8-10, 09:03 AM | #10 |
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A = [(ik + α)C + (α - ik)D] / 2α Is that what the book is looking for? |
| Sep8-10, 02:00 PM | #11 |
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Probably not. Try solving for C and D in terms of A. Then you can calculate the ratio C/D.
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| Sep8-10, 07:30 PM | #12 |
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Just now, I got A = [-2ik/(α-ik)] D. I suspect C would be something similar. |
| Sep8-10, 07:34 PM | #13 |
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I think that's right, and C comes out similarly.
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| Sep8-10, 07:42 PM | #14 |
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http://www.amazon.com/Quantum-Physic.../dp/0471057002 |
| Sep8-10, 07:51 PM | #15 |
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If you don't like that text, you might want to see if your library has Griffith's book on quantum mechanics. I'll admit I've never seen it, but I used his particle physics book as an undergrad. He was very good at explaining concepts and showing how to apply them in problems.
http://www.amazon.com/Introduction-Q...ref=pd_sim_b_1 (It's the top-selling book on quantum mechanics at Amazon.) |
| Sep8-10, 07:55 PM | #16 |
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| Sep8-10, 11:51 PM | #17 |
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Yes, I think it would help. It couldn't hurt. Textbooks are so expensive, though, so I'd try to look through a copy in a bookstore or at the library first.
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